Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.

WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).

I build up this question while thinking of the question Mappings preserving convex polyhedra.

share|cite|improve this question
You tagged it with "compactness" but didn't make any mention in the question, so I feel like some telepathy (or some subliminal messaging) might be going on here. Can you clarify any hypotheses you are thinking of? – rschwieb May 5 '14 at 20:07
@rschwieb It is useful to note that closed intervals are precisely the compact connected subsets of $\mathbb R$. – Alex Becker May 5 '14 at 20:09
@AlexBecker It's useful to know that's what the poster means by closed intervals, too, if that's the case. – rschwieb May 5 '14 at 20:10
@rschwieb, I had in mind that closed and bounded interval are the compact sets. Sorry for not mentioning it in the post. – Gilles Bonnet May 5 '14 at 20:12
@GillesBonnet Thanks! That makes everything clear. – rschwieb May 5 '14 at 20:13
up vote 10 down vote accepted

This is false. Consider $$f(x)=\begin{cases} \sin \frac{1}{x} & x\ne 0\\ 0 &x = 0 \end{cases}$$ which is discontinuous at $0$. However, for any closed interval $I$ not containing $0$, we have that $f(I)$ is a closed interval since $f$ is continuous on $I$, and for any closed interval $I$ containing $0$ we have $f(I)=[-1,1]$.

share|cite|improve this answer
ohh! classic! Wonder how I did not think about it myself! – Gilles Bonnet May 5 '14 at 20:14
@GillesBonnet What I noticed was that your property is very much like the Intermediate Value Property, and this function is famously discontinuous but satisfies IVP. – Alex Becker May 5 '14 at 20:16
@AlexBecker, $\frac{\sin x}{x} \rightarrow 1^-$ when $x\rightarrow 0$ so the image of $(-\varepsilon,\varepsilon)$ (for $\varepsilon$ small enough)its of the form $\{0\}\cup (1-\delta,1)$ tha it isnt a closed interval – themaker May 5 '14 at 20:18
ok, thanks. Meta question: should I edit the title to: "Map preserving intervals but discontinuous." in order to make it more clear. – Gilles Bonnet May 5 '14 at 20:19
@themaker You're right, I made a typo. Should be right now. – Alex Becker May 5 '14 at 20:19

A closed map* is not necessarily continuous. As mentioned in e.g., Wikipedia's page, the argument function assigning to a point p in $S^1$ its "total argument" (in the range $[0, 2\pi)$), i.e., the arc-length in radians at any x in $S^1$ , measured from a fixed point p, is both open, closed, but not continuous ; specifically near the fixed point $p$ , where the value goes from being close to $2\pi$ to being $0$.

share|cite|improve this answer
This property is stronger than being a closed map however. – Alex Becker May 5 '14 at 20:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.